C++11递归原子自旋锁
C++11递归原子自旋锁
提问于 2015-07-02 14:36:31
你能回顾一下我的自旋锁实现吗？
class spinlock {
private:
static const std::thread::id lock_is_free;
std::atomic<std::thread::id> lock_owner;
int lock_count;
public:
spinlock() : lock_owner(lock_is_free), lock_count(0) {
}
void lock() {
static thread_local std::thread::id this_thread_local = std::this_thread::get_id();
auto this_thread = this_thread_local;
for (auto id = lock_is_free; ; id = lock_is_free) {
if (lock_owner.compare_exchange_strong(id, this_thread, std::memory_order_acquire, std::memory_order_relaxed)) { ++lock_count; break; }
id = this_thread;
if (lock_owner.compare_exchange_strong(id, this_thread, std::memory_order_acquire, std::memory_order_relaxed)) { ++lock_count; break; }
}
}
void unlock() {
assert(lock_owner.load(std::memory_order_relaxed) == std::this_thread::get_id());
if (lock_count > 0 && --lock_count == 0) {
lock_owner.store(lock_is_free, std::memory_order_release);
}
}
};
const std::thread::id spinlock::lock_is_free;回答 1
Code Review用户
回答已采纳
发布于 2015-07-06 14:26:06
使用TLS
我还没有完全接受你对TLS的使用。对我来说，这看起来是一个不成熟的优化，也就是说，如果你已经测量了，并且它提高了你的性能，然后保持它。
静态成员
我认为没有这个静态成员会更好，但这是非常主观的。
可读性
你的代码在这里：
for (auto id = lock_is_free; ; id = lock_is_free) {
if (lock_owner.compare_exchange_strong(id, this_thread, std::memory_order_acquire, std::memory_order_relaxed)) { ++lock_count; break; }
id = this_thread;
if (lock_owner.compare_exchange_strong(id, this_thread, std::memory_order_acquire, std::memory_order_relaxed)) { ++lock_count; break; }
}对我来说很难读懂。事实上，我很难理解这种行为是什么。在我看来你真正想要的是：
// If we're being locked recursively, this will always compare false
// (insert your desired memory_order here) because no one can fiddle
// with `lock_owner` while we have the lock. If we're not entering
// recursively then this will always compare true because the thread
// doing the checking can only be in one place at one time.
if(lock_owner.load() != this_thread){
// Okay so it's not a recursive call.
do{
auto id = lock_is_free;
}while(!lock_owner.compare_exchange_weak(id, this_thread,
std::memory_order_acquire,
std::memory_order_relaxed));
}
// Okay in any event, recursive or not we have the lock now.
lock_count++;
return;注意:我将compare_exchange_strong更改为compare_exchange_weak，这是在循环中按Std：：原子：：比较_交换(cppreference.com)调用compare_exchange时推荐的。
我更喜欢这种unlock()的实现
void unlock(){
assert(lock_owner.load() == std::this_thread::get_id());
assert(lock_count != 0);
--lock_count;
if(lock_count == 0){
lock_owner.store(lock_is_free, std::memory_order_release);
}
}按照std::mutex::unlock()的样式，在调用线程不拥有锁时调用unlock是未定义的行为，我们可以简单地使用这两个断言来满足自己的要求，以便在调试时提供帮助。
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原文链接：
https://codereview.stackexchange.com/questions/95590
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